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There are 12,000 Btus per ton of cooling.
In order to determine tons of cooling, the formula is:
Tons = (dT x GPM) / 24
If you would like to know the tons of cooling in Btus, you would multiply this formula by 12,000..
Tons (Btu) = (dT x GPM x 12,000) / 24
Basic arithmetic yields:
Tons (Btu) = (dT x GPM x 500)
{12,000 / 24 = 500}
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How many BTU per Hour equals 1 kilowatt?
1kW is 3,412.13 BTU/hr
One kilowatt equals how many btu per hour?
A BTU is about 1055 joules. A kilowatt is 1000 joules/second, so it is 3,600,000 joules/hour. Dividing that by 1055 joules gives you the equivalent of about 3400 BTU/hour.
10000 btu equals how many pounds ao propane an hour?
A 10,000 BTU heater will use about 1/2 pound of propane per hour. Conversions to different types of energy being used (ie steam, KwH, etc.) will vary the answer, as will variations such as leaks, 02, temperature setting, efficiency, etc.
How many btu's would it take to raise the temperature 1 degree?
This question can not be answered without know much more information. Such as the material that needs to have its temperature changed. How much of that material there is.
How many btus to raise 1 pound of water from room temp deg to steam 213 deg?
First convert 1 lb of water to lb-moles which is 0.055 lb-moles (you'll need this later). This problem can be broken into 3 steps:(1) You need to detemine how much heat is needed to raise room temperature water (68oF) to 212oF. This can be used using the heat capacity of water which at room temperature is 1 Btu/lboF. So the amount of heat needed for this is:Q1 = m*Cp*ΔT= (1 lb)*(1 Btu/lboF)*(212 - 68oF)= 144 Btu(2) Next you need to account for the phase change. The water changes to steam at 212oF. You use the heat of vaporization which you can look up in any Chemistry or Chemical Engineering Handbook. The Hvap that I found is 17493.5 Btu/lb-mole.Q2 = n(lb-moles)*Hvap= (0.055 lb-moles)*(17493.5 Btu/lb-mole)= 972.64 Btu(3) Next you need to find out how much heat is needed to raise the temperature of the steam from 212 to 213oF. You can look up the heat capacity of steam at 212oF to be 0.485 Btu/lboF.Q3 = m*Cp*ΔT= (1 lb)*(0.485 Btu/lboF)*(213-212oF)= 0.485 BtuTo find the total heat needed add Q1+Q2+Q3 (144+972.64+0.485) =1117.12 Btu
